UDF解决方案效率低下（版本无关）：

from pyspark.sql.types import StringType
from pyspark.sql.functions import udf

def translate(mapping):
    def translate_(col):
        return mapping.get(col)
    return udf(translate_, StringType())

df = sc.parallelize([('DS', ), ('G', ), ('INVALID', )]).toDF(['key'])
mapping = {
    'A': 'S', 'B': 'S', 'C': 'S', 'DS': 'S', 'DNS': 'S', 
    'E': 'NS', 'F': 'NS', 'G': 'NS', 'H': 'NS'}

df.withColumn("value", translate(mapping)("key"))

结果是：

+-------+-----+
|    key|value|
+-------+-----+
|     DS|    S|
|      G|   NS|
|INVALID| null|
+-------+-----+

更高效的方法（Spark〉= 2.0，Spark〈3.0）是创建一个MapType常量：

from pyspark.sql.functions import col, create_map, lit
from itertools import chain

mapping_expr = create_map([lit(x) for x in chain(*mapping.items())])

df.withColumn("value", mapping_expr.getItem(col("key")))

结果是一样的

+-------+-----+
|    key|value|
+-------+-----+
|     DS|    S|
|      G|   NS|
|INVALID| null|
+-------+-----+

但更高效的执行计划：

== Physical Plan ==
*Project [key#15, keys: [B,DNS,DS,F,E,H,C,G,A], values: [S,S,S,NS,NS,NS,S,NS,S][key#15] AS value#53]
+- Scan ExistingRDD[key#15]

与UDF版本相比：

== Physical Plan ==
*Project [key#15, pythonUDF0#61 AS value#57]
+- BatchEvalPython [translate_(key#15)], [key#15, pythonUDF0#61]
   +- Scan ExistingRDD[key#15]

在Spark〉= 3.0中，getItem应替换为__getitem__（[]），即：

from pyspark.sql.functions import col, create_map, lit
from itertools import chain

mapping_expr = create_map([lit(x) for x in chain(*mapping.items())])

df.withColumn("value", mapping_expr[col("key")])

听起来最简单的解决方案是使用replace函数：http://spark.apache.org/docs/2.4.0/api/python/pyspark.sql.html#pyspark.sql.DataFrame.replace

mapping= {
        'A': '1',
        'B': '2'
    }
df2 = df.replace(to_replace=mapping, subset=['yourColName'])

如果你想从嵌套字典创建一个Map列，你可以使用：

def create_map(d,):
    if type(d) != dict:
        return F.lit(d)

    level_map = []
    for k in d:
        level_map.append(F.lit(k))
        level_map.append(create_map(d[k]))
    return F.create_map(level_map)

d = {'a': 1, 'b': {'c': 2, 'd': 'blah'}}
print(create_map(d)) # <- Column<b'map(a, 1, b, map(c, 2, d, blah))'>

如果没有itertools导入，列表解析可以很好地处理它。

来自法令的Map：

F.create_map([F.lit(x) for i in dic.items() for x in i])

提取值：

F.create_map([F.lit(x) for i in dic.items() for x in i])[F.col('col1')]

全检：

from pyspark.sql import functions as F
df = spark.createDataFrame(
    [('A',), ('E',), ('INVALID',)],
    ['col1']
)
dic = {'A': 'S', 'B': 'S', 'E': 'NS'}

map_col = F.create_map([F.lit(x) for i in dic.items() for x in i])
df = df.withColumn('col2', map_col[F.col('col1')])

df.show()
# +-------+----+
# |   col1|col2|
# +-------+----+
# |      A|   S|
# |      E|  NS|
# |INVALID|null|
# +-------+----+

在Spark SQL中可以使用字典转换为大小写语法的函数

func_mapper = lambda dic,col,default : f"(CASE {col} WHEN " + " WHEN ".join([ f"'{k}' THEN '{v}'" for (k,v) in dic.items() ]) + f" ELSE '{default}' END)"

如果有人也需要Map空值，the accepted answer不适合我，map类型的问题是它不能处理空值键。
但是我们可以用生成的CASE WHEN语句替换它，并使用isNull代替== None：

from pyspark.sql import functions as F
from functools import reduce

d = spark.sparkContext.parallelize([('A', ), ('B', ), (None, ), ('INVALID', )]).toDF(['key'])

mapping = {'A': '1', 'B': '2', None: 'empty'}
map_tuples = list(mapping.items())

def email_eq_null_safe(key):
    if key is None:
        return F.col('key').isNull()
    else:
        return F.col('key') == key

'''
F.when(
    F.col('key') == key1,
    value1
).when(
    F.col('key') == key2,
    value2
)....
'''
whens = reduce(
    lambda prev, nxt: prev.when(email_eq_null_safe(nxt[0]), nxt[1]), 
    map_tuples[1:],
    F.when(email_eq_null_safe(map_tuples[0][0]), map_tuples[0][1])
)

d.select(
    'key',
    whens.alias('value')
).show()

+-------+-----+
|    key|value|
+-------+-----+
|      A|    1|
|      B|    2|
|   null|empty|
|INVALID| null|
+-------+-----+